package com.sxkiler.demo.easy;

import org.junit.jupiter.api.Assertions;
import org.junit.jupiter.api.Test;
import java.util.*;
import com.sxkiler.demo.model.*;

/**
range-sum-of-bst=二叉搜索树的范围和
<p>给定二叉搜索树的根结点&nbsp;<code>root</code>，返回 <code>L</code> 和 <code>R</code>（含）之间的所有结点的值的和。</p>

<p>二叉搜索树保证具有唯一的值。</p>

<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>

<pre><strong>输入：</strong>root = [10,5,15,3,7,null,18], L = 7, R = 15
<strong>输出：</strong>32
</pre>

<p><strong>示例&nbsp;2：</strong></p>

<pre><strong>输入：</strong>root = [10,5,15,3,7,13,18,1,null,6], L = 6, R = 10
<strong>输出：</strong>23
</pre>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ol>
	<li>树中的结点数量最多为&nbsp;<code>10000</code>&nbsp;个。</li>
	<li>最终的答案保证小于&nbsp;<code>2^31</code>。</li>
</ol>

 */
public class rangeSumBST {
    

    class Solution {
        public Integer rangeSumBST(TreeNode param0,Integer param1,Integer param2) {
            return null;
        }
    }

    @Test
    public void test(){
        Solution solution = new Solution();
        /**
        [10,5,15,3,7,null,18]
7
15
        */
        //int [] num1 = new int[]{1,3};
        //int [] num2 = new int[]{2};
        //Assertions.assertEquals(solution.{{questionName}}(num1,num2),2);
    }
}

